[RC5] RC5 question
Bruce Wilson
bruce.wilson at distributed.net
Fri Oct 3 21:58:32 EDT 2003
Cool numbers. I can live with an initial estimate of 101 years.
What you also haven't factored in is the possibility of enlisting more
people than we currently have. Of course, some may choose to work on
OGR or The Next Big Thing (whatever that may be). Still, we can
exceed these estimates and finish sooner than your estimate by
attracting more people to participate.
__
Bruce Wilson <bwilson at distributed.net>
PGP KeyID: 5430B995, http://www.toomuchblue.com/
"I want to move to Theory. Everything works in Theory."
John Cash, id Software
 Original Message
 From: rc5bounces at lists.distributed.net
 [mailto:rc5bounces at lists.distributed.net] On Behalf Of
 Daniel Quintiliani
 Sent: Friday, October 03, 2003 20:33
 To: D.net Discussion
 Subject: Re: [RC5] RC5 question


 On Sat, 4 Oct 2003 09:05:04 +0800, Elektron
 <elektron_rc5 at yahoo.ca> wrote:

 >
 > Taking the latest RC564 numbers,
 > 10,428,970,063,364,096 Keys were completed yesterday
 (0.056536% of the
 > keyspace) at a sustained rate of 120,705,672,030 Keys/sec.
 >
 > That's 3809103098378452560 Keys/year (let's call it k). If
 computing
 > power doubles every 18 months (1.5 years), then a^1.5=2, or
 a=2^(1/1.5)
 > or approx. 1.587401051968. So our keyrate with respect to
 time t (years)
 > is ka^t, so our keys tested (big K) (by integration) is
 ka^t ln(a) + C,
 > and since we have no keys tested when t=0, k ln(a) + C = 0,
 so K = ka^t
 > ln(a)  k ln(a). K=2^128.
 >
 > 2^128 = k ln(a) (a^t1)
 > 2^128/(k ln(a)) = a^t1
 > 2^128/(k ln(a)) + 1 = a^t
 > t = ln(2^128/(k ln(a)))/ln(a) = 101.08435774324143022752
 >
 > Of course, I'm assuming RC5128 is as fast as RC564 (which
 it isn't),
 > and that moore's law holds (which it might). Interestingly,
 the last time
 > I did this, I got 600 years. Somebody hit me if I did my
 math wrong. But
 > either way, 101 years is a lot more than my lifetime.
 >

 Using yesterday's RC572 rate for a more accurate measure of time:

 114327804047 Keys/s * 60 * 60 * 24 * 365.25
 (goes upstairs and grabs TI83+ from backpack)
 k = 3.607911127 * 10^8

 Plugging it into your formula:

 t = ln(2^128/((3.607911127*10^8) ln(2^(1/1.5)))) / ln(2^(1/1.5))

 t = 101.201789 years

 
 Dan
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